Concept of Hardy-Weinberg Equilibrium Explained

Before we start the Hardy-Weinberg equilibrium, let’s talk about the meaning of evolution. Does evolution mean the development of new species? Well, the answer is absolutely no. A slight change within a species from one generation to another over a long period of time can result in the gradual transition of new species.

Biological sciences define evolution as the sum total of all the genetically inherited changes in individuals who are the members of a population’s gene pool (the sum total of all the alleles in a population). It is clear that the effects of evolution are experienced by individuals, but it is the overall population that actually evolves. So, we can define evolution as the change in frequencies of alleles in the gene pool of a population.

In the early 20^th century, English mathematician Godfrey Hardy and the German physician Wilhelm Weinberg independently provided the explanation of concept called Hardy-Weinberg Equilibrium. This theory states that a population’s allele and genotype frequencies are inherently stable unless some kind of evolutionary force is acting on the population, the population would carry the same alleles in the same populations generation after generation. Individuals in a population would look essentially the same and this would be unrelated to whether the alleles are dominant or recessive. However, this concept is true only when the five evolutionary forces are absent as these forces can disrupt the equilibrium. These five forces include:

• Natural Selection
• Mutation
• Genetic Drift
• Gene migration or gene flow
• Genetic recombination

It means, if the above five factors are absent in a population, no evolution will occur in that particular population. However, in a practical sense, it is not possible that any of these factors would not be present in nature. In other words, we can say that evolution is the inevitable result.

Hardy and Weinberg developed a simple equation that can be used to find the probable genotype frequencies in a population and changes if any can be tracked from one generation to another. This is known as the Hardy-Weinberg Equilibrium equation. 

P² + 2pq + q² = 1          or, (p + q)² = 1

Where p represents dominant allele A; q represents recessive allele a

In other words, p = AA + ½ Aa;     q = aa + ½ Aa    or p + q = 1

In the above equation, p² is the predicted frequency of homozygous dominant (AA) people in a population; 2pq is the predicted frequency of heterozygous (Aa) people and q² is the predicted frequency of homozygous recessive (aa) people.

It is interesting to note that only the frequency of homozygous recessive people can be calculated. As far as a dominant trait is considered, it is represented by either homozygous dominant (p²) or heterozygous (2pq). However, using the Hardy-Weinberg equation, we can also calculate the frequency of p² and 2pq without any hassle. Since, p + q = 1 and p = 1-q where q is known, so we can calculate the frequency of p² and 2pq easily. 

Q1. In a given population, the percentage of homozygous recessive genotype (aa) is 36%. What would be the frequency of:

a)      a allele

b)      A allele

c)       AA and Aa genotype

Solution: Since the percentage of recessive genotype is given (36%), it means q² = 36% = 0.36

So, q = 0.6, and since q= a; the frequency of a = 60%

The frequency of A can be calculated from the equation p + q = 1

P = 1 – q

P = 1 – 0.6 = 0.4; so the frequency of A is 40%

The frequency of genotype AA and Aa

The frequency of AA equals to p²  (0.4x0.4 = 0.16) = 16%

The frequency of Aa equals 2pq = (2 x 0.4 x 0.6) = 0.48 = 48%

Q2. Sickle-cell anemia is a genetic disease. Normal homozygous individuals (SS) having normal red blood cells are susceptible to the malaria parasite. On the other hand, individuals with sickle-cell trait (ss) have red blood cells that easily collapse under deoxygenated condition. These individuals are resistant to malaria parasite as such parasites cannot grow in the sickle-celled red blood cells. Individuals with sickle-cell red blood cells often die because of the genetic defect. However, individuals with heterozygous condition (Ss) have some sickle-celled red blood cells, but not enough to cause mortality. Besides, malaria parasite can also survive within these partially defective red blood cells. In other words, a heterozygous condition is better than either of the homozygous conditions. If 9% of an African population is born with sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

Ans: ss = q² = 9% = 0.09; q = 0.3

P = 1 – q = 1 – 0.3 = 0.7

2pq = 2 x 0.7 x 0.3 = 0.42 = 42 % (heterozygous carrier)

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What is Chargaff's Rule of DNA?

Erwin Chargaff was an Australian biochemist who analyzed nucleic acids of many organisms. From his research, he concluded that the number of purine bases (A, G) equals the pyrimidine bases (C, T) in any double stranded DNA of an organism.

It means,

•  A+G / T+C = 1; 
• A+T / C+G may vary from organism to organism.
• % A = % T, and % G = % C

Based on Chargaff's rule, many questions are asked in various competitive exams including AIPMT, CSIR NET, GATE, etc.

Q1. One of the strands of a double stranded DNA has the following number of bases:
A= 3800; T= 2600. What would be the base composition of the double stranded DNA?

Solution:

From the question, it is a double stranded DNA (Chargaff's rule is valid for only double stranded DNA), and only one strand with its base composition is given. So, the total number of bases would be 20k (10k bases on each strand).

Now, A and T bases with their numbers are given on one strand, so A will have T and T will have A on their complementary DNA. It means, on one strand, the total number of A would be 3800+2600 because % A = % T. Similarly, the total number of T would be 2600 + 3800 = 6400

The total number of A and T on both strands would be 6400 + 6400 = 12800

The remaining bases would be G and C = 20,000 - 12800 = 7200

Since % G =  % C; G = 7200/2 = 3600; C = 3600

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How to Analyze Autosomal Recessive Pedigree Trait?

In order to analyze this pedigree, we have to first check whether it is dominant or recessive in nature.

The criteria for dominant pedigree trait is:

a. There should not be skipping of generation
b. Either or both the parents should be affected to pass a disease to the next generation. 

Note: when two normal parents transfer a disease to the offspring, it is a recessive characteristic.

Now, when you see the above pedigree, you can see that in the second generation, two parents 8 and 9 are normal, but passing a disease to their daughter (17) in the third generation. So, you can say that the pedigree is recessive.

The second step is to check whether it is autosomal recessive or X-linked recessive. The best way is to first check for X-linked. If it is X-linked, it should follow two conditions:

a. More boys affected than girls
b. Affected mom passes the disease to all her sons

You can see that in the first generation, an affected mom does not pass disease to all her sons, so, it is not an X-linked trait.

So, we can conclude that the above pedigree shows autosomal recessive trait.

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